A secant line intersects the curve $g(x)=e^x$ at two points with $x$ -coordinates $1$ and $1+h$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{e^{1+h}-e^}{1+h}$ (Choice B) B $\dfrac{e^{1+h}-e^}{1}$ (Choice C) C $\dfrac{e^{ h}-e^}{h}$ (Choice D) D $\dfrac{e^{1+h}-e^}{h}$
Answer: We are given that the secant line intersects the curve at $x=1$ and $x=1+h$. Since these points are on the graph of $g(x)=e^x$, we know that they must be $(1, e)$ and $(1+h, e^{{1+h}})$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{e^{{1+h}}-e}{1+h-1} \\\\ &=\dfrac{e^{{1+h}}-e}{h} \end{aligned}$ In conclusion, the slope of the secant line is $\dfrac{e^{{1+h}}-e}{h}$.